ĐK: \(x,y\neq 0\)
\(PT\left(2\right)\Leftrightarrow x=9-y\)
Thay vào \(PT\left(1\right)\Leftrightarrow\dfrac{1}{9-y}+\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow2y+18-2y=9y-y^2\)
\(\Leftrightarrow y^2-9y+18=0\\ \Leftrightarrow\left[{}\begin{matrix}y=3\Rightarrow x=6\\y=6\Rightarrow x=3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;3\right);\left(3;6\right)\)
ĐK x;y \(\ne\)0
HPT <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=18\\x+y=9\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x\left(9-x\right)=18\\y=9-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+18=0\\y=9-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x-6\right)=0\\y=9-x\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\\y=9-x\end{matrix}\right.\)
Khi x= 3 => y = 6
Khi x = 6 => y = 3
Vậy nghiệm phương trình là (3;6) ; (6;3)
